Bob is training for a fitness competition. In order to increase his maximum number of pull-ups, he follows the following routine: he begins with 25 pull-ups, rests for thirty seconds, and then does 24 pull-ups and rests, dropping one pull-up each time (25, 24, 23, etc.) until his final set of 11 pull-ups. How many total pull-ups does Bob do?
(A) 55
(B) 150
(C) 270
(D) 275
(E) 325
The OA is C.
I can do it making the whole sum: 25+24+23+22+...+11, but, is there another way (faster or easier) to make it?
Experts, can you give me a hand here?
Bob is training for a fitness competition. In order to. . .
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Great question - there is a faster way! When we have any group of terms, we can find the sum of all of those terms with:
$$mean\ \cdot\ number\ of\ terms$$
When we have a group of consecutive terms, we can find the mean of those terms with:
$$\frac{first\ term+last\ term}{2}$$
Combining those two equations, we can find the sum of a group of consecutive terms with:
$$\frac{first\ term+last\ term}{2}\cdot\ number\ of\ terms$$
Here, this gives:
$$\frac{25\ +\ 11}{2}\ \cdot\ 15\ =\ 270$$
$$mean\ \cdot\ number\ of\ terms$$
When we have a group of consecutive terms, we can find the mean of those terms with:
$$\frac{first\ term+last\ term}{2}$$
Combining those two equations, we can find the sum of a group of consecutive terms with:
$$\frac{first\ term+last\ term}{2}\cdot\ number\ of\ terms$$
Here, this gives:
$$\frac{25\ +\ 11}{2}\ \cdot\ 15\ =\ 270$$
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M7MBA wrote:Bob is training for a fitness competition. In order to increase his maximum number of pull-ups, he follows the following routine: he begins with 25 pull-ups, rests for thirty seconds, and then does 24 pull-ups and rests, dropping one pull-up each time (25, 24, 23, etc.) until his final set of 11 pull-ups. How many total pull-ups does Bob do?
(A) 55
(B) 150
(C) 270
(D) 275
(E) 325
The OA is C.
We need to calculate the sum of the consecutive integers from 11 to 25, inclusive. Rather than calculate 11 + 12 + 13 + ... + 24 + 25, we can calculate the average of the integers and then multiply the average by the number of integers to get the sum.
We know that there are (25 - 11) + 1 = 14 + 1 = 15 integers.
To calculate the average of the 15 integers, we use (smallest + largest)/2 = (11 + 25)/2 = 18.
The average is 18. Thus, the sum is 18 x 15 = 270.
Answer: C
I can do it making the whole sum: 25+24+23+22+...+11, but, is there another way (faster or easier) to make it?
Experts, can you give me a hand here?
Scott Woodbury-Stewart
Founder and CEO
[email protected]
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